Braking distance

The car travels at an average speed of 12 km/h and detects an obstacle 10 m in front of it. At 1 m in front of the obstacle it already runs 2 km/h. What is the braking distance? What is the required deceleration for stop in:

A) 1m
B) 1s?

Correct result:

x =  12.96 m
a2 =  0.5556 m/s2

Solution:

v1=12 km/h m/s=12/3.6  m/s=3.33333 m/s s1=10 m s2=1 m v2=2 km/h m/s=2/3.6  m/s=0.55556 m/s  s=s1s2=101=9 m s=12at2 at=(v1v2)  s=12(v1v2) t  t=2 s/(v1v2)=2 9/(3.33330.5556)=16225=6.48 s  a=v1v2t=3.33330.55566.48=62514580.4287 m/s2  t1=v1/a=3.3333/0.4287=972125=7.776 s  x=12 a t12=12 0.4287 7.7762=32425=32425 m=12.96 m
t2=2 s1/(v10)=2 10/(3.33330)=6 s a2=v1t2=3.33336=59=0.5556 m/s2



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